UNIFORMLY ERGODIC PROBABILITY MEASURES

Abstract

Let $G$ be a locally compact group and $\mu$ be a probability measure on $G$. We consider the convolution operator ${\lambda }_1\left(\mu \right):L_1\left(G\right)\to L_1\left(G\right)$ given by ${\lambda }_1\left(\mu \right)f=\mu *f$ and its restriction ${\lambda }_1^0\left(\mu \right)$ to the augmentation ideal $L_1^0\left(G\right)$. Say that $\mu$ is uniformly ergodic if the Cesàro means of the operator ${\lambda }_1^0\left(\mu \right)$ converge uniformly to $0$, that is, if ${\lambda }_1^0\left(\mu \right)$ is a uniformly mean ergodic operator with limit $0$, and that $\mu$ is uniformly completely mixing if the powers of the operator ${\lambda }_1^0\left(\mu \right)$ converge uniformly to $0$.

We completely characterize the uniform mean ergodicity of the operator ${\lambda }_1\left(\mu \right)$ and the uniform convergence of its powers, and see that there is no difference between ${\lambda }_1\left(\mu \right)$ and ${\lambda }_1^0\left(\mu \right)$ in these regards. We prove in particular that $\mu$ is uniformly ergodic if and only if $G$ is compact, $\mu$ is adapted (its support is not contained in a proper closed subgroup of $G$), and $1$ is an isolated point of the spectrum of $\mu$. The last of these three conditions can actually be replaced by $\mu$ being spread out (some convolution power of $\mu$ is not singular). The measure $\mu$ is uniformly completely mixing if and only if $G$ is compact, $\mu$ is spread out, and the only unimodular value in the spectrum of $\mu$ is $1$.